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Find lost element from a duplicated array

Given two arrays which are duplicates of each other except one element, that is one element from one of the array is missing, we need to find that missing element.

Examples:

Input:  arr1[] = {4, 1, 5, 9, 7}         arr2[] = {4, 5, 9, 7} Output: 1 1 is missing from second array.  Input: arr1[] = {2, 3, 4, 5}        arr2[] = {2, 3, 4, 5, 6} Output: 6 1 is missing from first array.

We strongly recommend you to minimize your browser and try this yourself first.

One simple solution is to iterate over arrays and check element by element and flag the missing element when an unmatch is found, but this solution requires linear time over size of array.

Another efficient solution is based on binary search approach. Algorithm steps are as follows:

  1. Start binary search in bigger array and get mid as (lo + hi) / 2
  2. If value from both array is same then missing element must be in right part so set lo as mid
  3. Else set hi as mid because missing element must be in left part of bigger array if mid elements are not equal.
  4. Special case are handled separately as for single element and zero element array, single element itself will be the missing element.

If first element itself is not equal then that element will be the missing element.

Below is C++ implementation of above steps

// C++ program to find missing element from same // sorted arrays (except missing element) #include <bits/stdc++.h> using namespace std;  // Funtion to find missing element based on binary // search approach.  arr1[] is of larger size and // N is size of it.  arr1[] and arr2[] are assumed // to be sorted arrays. int findMissingUtil(int arr1[], int arr2[], int N) {     // special case, for only element which is     // missing in second array     if (N == 1)         return arr1[0];      // special case, for first element missing     if (arr1[0] != arr2[0])         return arr1[0];      // Initialize current corner points     int lo = 0,  hi = N - 1;      // loop until lo < hi     while (lo < hi)     {         int mid = (lo + hi) / 2;          // If element at mid indices are equal         // then go to right subarray         if (arr1[mid] == arr2[mid])             lo = mid;         else             hi = mid;          // if lo, hi becomes contiguous,  break         if (lo == hi - 1)             break;     }      // missing element will be at hi index of     // bigger array     return arr1[hi]; }  // This function mainly does basic error checking // and calls findMissingUtil void findMissing(int arr1[], int arr2[], int M, int N) {     if (M = N-1)         cout << "Missing Element is "             << findMissingUtil(arr1, arr2, M) << endl;     else if (N == M-1)         cout << "Missing Element is "              << findMissingUtil(arr2, arr1, N) << endl;     else         cout << "Invalid Input"; }  // Driver Code int main() {     int arr1[] = {4, 1, 5, 9, 7};     int arr2[] = {4, 5, 9, 7};      int M = sizeof(arr1) / sizeof(int);     int N = sizeof(arr2) / sizeof(int);      findMissing(arr1, arr2, M, N);      return 0; }

Output :

Missing Element is 1

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